This is so far my understanding of classical mechanics, Interspaced are a few questions where I am still not entirely sure what is going on. Thank you for your help !!
A tangent space can be thought of as the space of all possible tangent vectors to a point $p$ on an arbitrary manifold $M$, $M$ is the base space for all the following geometrical structures i.e vector spaces, tangent spaces, fibre bundles etc.
1) Is it the base space for all additional structures?
A tangent bundle is a fibre bundle consisting of all the tangent spaces for all $p\in M$, (so I'm guessing its dimensionality is huge?). A vector field is a space where one of the vectors from each tangent space is selected, its dimensionality is twice that of $M$?
2) What is the object/equation that chooses which vector out of the tangent space is selected for each point?. Is this what the $\textit{projection}$ does?
A one form is a function that maps a point in a vector space defined on a cotangent bundle, to a real number.
3) What does it mean that "a vector field on a cotangent fibre bundle has an associated one form"?
If we have some physical scalar quantity of a system we describe it by a vector field on $T^*M$, the cotangent bundle. The associated one forms to this vector field are the gradients of the quantities of the physical properties of the system. How does this tie in with the definition above of a one form?
4) If this is so why then do we need a two form for phase space?
In the OP a lot of different questions are stated. I try to discuss them and introduce some definitions. If $M$ is a smooth real manifold of finite dimension $n$, then
I do not fully understand the question. The sentence "A vector field is a space where one of the vectors from each tangent space is selected, its dimensionality is twice that of M" is not clear to me; let us discuss it. A vector field $\Phi$ on $M$ is a smooth map $\Phi: M\rightarrow TM$, s.t. $\Phi(x)=(x,\Phi_x)$ and $\Phi_x\in T_x M$ for all $x\in M$. As the dimension of $T_xM$ is equal to the dimension of $M$, for all $x\in M$, then $\Phi_x$ is uniquely written as a linear combination of basis elements in $T_xM$ (again for all $x\in M$). More clearly: the tangent bundle $TM$ has no linear space structure (by definition); each tangent space $T_xM$ has a linear space structure of dimension $n$, instead. On the other hand, the tangent bundle $TM$ is naturally endowed with a smooth manifold structure s.t. its dimension (as smooth manifold) is equal to $2n$.
No, the projection $\pi:TM\rightarrow M$ does not what is stated in the question; in fact $\pi(x, v):= x$ for all $x\in M$; in other words, the projection "forgets" pointwise the tangent space structure projecting on the base manifold. The sentence on the one forms makes no sense to me; a one form is a (smooth) section of the cotangent bundle $T^*M$; for all $x\in M$ the restriction $\omega_x$ of the one form $\omega$ to $x$ is a linear functional on the tangent space $T_x M$. Please check this link for more details.
In this case I refer to this page for the definitions of dual space and linear functionals. You need to apply those considerations in the case $V:=T_x M$ and $V^*:= T^*_x M$, with $x\in M$.
The two form on the phase space you are referring to is a symplectic 2-form; this is an extra structure on a manifold, which is therefore called symplectic manifold. The topic is very wide and rich: I refer to these notes (in particular to pag. 6) for an introduction to the topic.